All categories

3 years ago

HELP ME PLEASE !

Mathematics

1 answer:

STatiana [176]3 years ago

8 0

1.+-7, 2.+-8, 3.12, 5.22

You might be interested in

Brian invests £4000 into his bank account.

GuDViN [60]

Answer:

10.24 years

Step-by-step explanation:

The computation of the time period is as follows:

As we know that

Amount = Principal × (1 + rate of interest)^time period

£8000 = £4000 × (1 + 0.07)^time period

£8,000 ÷ £4,000 = 1.07^time period

2 = 1.07^time period

Now apply the log to the both sides

log 2 ÷ log 1.07 = time period

= 10.24 years

7 0

3 years ago

Divide using a common factor of 2 to find an equivalent fraction for 2/6

Ludmilka [50]

Answer:

4/12

Step-by-step explanation:

5 0

2 years ago

an ellipse has a center at the origin , a vertex along the major axis at (13,0), and a focus at (12,0). What is the equation of

-BARSIC- [3]

Answer:

The answer to your question is below

Step-by-step explanation:

Data

Center = (0, 0)

Vertex = (13, 0)

Focus = (12, 0)

Process

From the data we know that it is a horizontal ellipse.

1.- Calculate "a", the distance from the center to the vertex.

a = 13

2.- Calculate "c", the distance from the center to the focus

c = 12

3.- Calculate b

Use the Pythagorean theorem to find it

a² = b² + c²

-Solve for b

b² = a² - c²

-Substitution

b² = 13² - 12²

-Simplification

b² = 169 - 144

b² = 25

b = 5

4.- Find the equation of the ellipse

$\frac{x^{2} }{13^{2}} + \frac{y^{2}}{5^{2}} = 1$ or $\frac{x^{2} }{169} + \frac{y^{2}}{25} = 1$

7 0

3 years ago

If f(x) = x ^ 2 + 3 , then f(x+y)=?

kobusy [5.1K]

Answer:

f(x)=x^2+3

f(x+y)=(x+y)^2+3

this is done by replacing x with (x+y)

f(x+y)=x^2+2xy+y^2+3

therefore : f(x+y)= x^2+y^2+2xy+3

5 0

3 years ago

Use the power series for 1 1−x to find a power series representation of f(x) = ln(1−x). What is the radius of convergence? (Note

Viktor [21]

a. Recall that

$\displaystyle\int\frac{\mathrm dx}{1-x}=-\ln|1-x|+C$

For $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

By integrating both sides, we get

$\displaystyle-\ln(1-x)=C+\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

If $x=0$ , then

$\displaystyle-\ln1=C+\sum_{n=0}^\infty\frac{0^{n+1}}{n+1}\implies 0=C+0\implies C=0$

so that

$\displaystyle\ln(1-x)=-\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}$

We can shift the index to simplify the sum slightly.

$\displaystyle\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$

b. The power series for $x\ln(1-x)$ can be obtained simply by multiplying both sides of the series above by $x$ .

$\displaystyle x\ln(1-x)=-\sum_{n=1}^\infty\frac{x^{n+1}}n$

c. We have

$\ln2=-\dfrac\ln12=-\ln\left(1-\dfrac12\right)$

$\displaystyle\implies\ln2=\sum_{n=1}^\infty\frac1{n2^n}$

4 0

3 years ago